Commutative Algebra 46

Properties of Dedekind Domains

Throughout this article, A denotes a Dedekind domain.

Proposition 1.

Every fractional ideal of A can be written as

\mathfrak m_1^{k_1} \ldots \mathfrak m_n^{k_n}, \quad k_1, \ldots, k_n \in \mathbb Z - \{0\}

where each \mathfrak m_i is a maximal ideal. The expression is unique up to permutation of terms.


In particular, the maximal ideals of A generate its Picard group.


Existence. First suppose \mathfrak a \subseteq A is a non-zero ideal. If \mathfrak a = (1) we are done. Otherwise it is contained in some maximal ideal \mathfrak m, necessarily invertible. Now \mathfrak a \ne \mathfrak a\mathfrak m^{-1}, because if equality holds multiplying both sides by \mathfrak {ma}^{-1} gives \mathfrak m= A, so \mathfrak a \subsetneq \mathfrak a\mathfrak m^{-1} \subseteq A. We replace \mathfrak a by \mathfrak a \mathfrak m^{-1} and repeat the process; this cannot continue indefinitely or we would have

\mathfrak a \subsetneq \mathfrak a \mathfrak m_1^{-1}\subsetneq \mathfrak a \mathfrak m_1^{-1} \mathfrak m_2^{-1} \subsetneq \ldots

Hence \mathfrak a is a finite product of maximal ideals. Now a general fractional ideal is of the form \frac 1 x \mathfrak a for a non-zero ideal \mathfrak a \subseteq A and x\in A-\{0\}. Write both \mathfrak a and xA as a product of maximal ideals and we are done.

Uniqueness. Suppose \mathfrak m_1^{k_1} \ldots \mathfrak m_n^{k_n} = \mathfrak m_1^{j_1} \ldots \mathfrak m_n^{j_n}, where j_i, k_i \in \mathbb Z (some possibly zero). If (k_1, \ldots, k_n) \ne (j_1, \ldots, j_n), after cancelling terms and reordering, we obtain a relation of the form:

\mathfrak m_1^{k_1} \ldots \mathfrak m_s^{k_s} = \mathfrak m_{s+1}^{k_{s+1}} \ldots \mathfrak m_t^{k_t}, \quad k_i \in \mathbb Z_{>0}.

This is an equality of ideals. We get a contradiction because the LHS is contained in \mathfrak m_1 but the RHS is not. ♦

Exercise A

Prove that an integral domain is a PID if and only if it is a Dedekind domain and a UFD. Hence we have the following Venn diagram.


Now we have the following.

Proposition 2.

Let M, N be non-zero fractional ideals of A. Write, as in proposition 1,

M = \mathfrak m_1^{d_1} \mathfrak m_2^{d_2} \ldots \mathfrak m_k^{d_k}, \quad N = \mathfrak m_1^{e_1} \mathfrak m_2^{e_2} \ldots \mathfrak m_k^{e_k}

where each \mathfrak m_i \subset A is maximal and d_i, e_i \in \mathbb Z. Then

\begin{aligned} M+N &= \mathfrak m_1^{\min(d_1, e_1)} \ldots \mathfrak m_k^{\min(d_k, e_k)}\\ M\cap N &= \mathfrak m_1^{\max(d_1, e_1)} \ldots \mathfrak m_k^{\max(d_k, e_k)}\\ MN &= \mathfrak m_1^{d_1 + e_1} \ldots \mathfrak m_k^{d_k + e_k},\\ (M:N) &= \mathfrak m_1^{d_1 - e_1} \ldots \mathfrak m_k^{d_k - e_k}\end{aligned}


Exercise. [ Localize at each \mathfrak m_i; reduce to the case where A is a dvr. ]


Constructing Dedekind Domains

How do we find Dedekind domains? For starters, a PID is a Dedekind domain so we have \mathbb Z and k[X] (for any field k), but these are not too interesting.

Here is a standard way to construct new Dedekind domains from an existing one A. Let K be its field of fractions and L be a finite extension of K; we let B be the integral closure of A in L.


Exercise B

Prove that L is the field of fractions of B.

Now B has the following properties:

  • it is a normal domain by construction;
  • dim B = 1 since B is an integral extension of A, and dim A = 1 (by main theorem here).

Hence, if we can show that B is noetherian, then it is a Dedekind domain. But this result, called the Krull-Akizuki theorem, is surprisingly hard to prove. In most cases of interest, B will turn out to be a finitely generated A-module; thus B is noetherian as an A-module and hence as a module over itself.


1. Let K/\mathbb Q be a finite extension and \mathcal O_K be the integral closure of \mathbb Z in K. One can show that \mathcal O_K is a discrete subgroup of K, when we topologize K by identifying it with \mathbb Q^n. Hence \mathcal O_K is a finitely generated \mathbb Z-module; in particular it is a noetherian ring and hence a Dedekind domain. A more general result of this nature will be proven later.

2. Let k be any field such that \mathrm{char} k \ne 2. The integral closure of k[X] in L = k(X)[Y]/(Y^2 - X^3 + X) is A = k[X, Y]/(Y^2 - X^3 +X). [ Proof: exercise. ] Since A is clearly noetherian, it is a Dedekind domain.

Exercise C

Let m\in \mathbb Z - \{\pm 1\} be a square-free integer and K = \mathbb Q(\sqrt m), which is quadratic over \mathbb Q. Prove that

\mathcal O_K = \begin{cases} \mathbb Z[\frac{1 + \sqrt{m}}2], \quad &\text{if } m\equiv 1 \pmod 4, \\ \mathbb Z[\sqrt{m}], \quad &\text{if } m\equiv 2,3 \pmod 4.\end{cases}




Let K = \mathrm{Frac} A. Each maximal \mathfrak m \subset A gives a function

\nu_{\mathfrak m} : K - \{0\} \longrightarrow \mathbb Z,

which takes x to the exponent of \mathfrak m in the above factorization of xA. This is called the valuation of x at \mathfrak m.

Exercise D

1. Prove the following properties of \nu = \nu_{\mathfrak m}: for any x,y\in K - \{0\} we have:

\nu(1) = 0, \quad \nu(xy) = \nu(x) + \nu(y), \quad \nu(x+y) \ge \min(\nu(x), \nu(y)) \text{ if } x+y\ne 0.

Verify that if we set \nu(0) = \infty these properties hold for all x,y\in K.

2. Define a distance function on K by d(x, y) = e^{-\nu(x-y)}. Prove that (Kd) is a metric space, where the metric satisfies the strong triangular inequality.

x,y, z\in K \implies d(x, z) \le \max(d(x, y), d(y, z)).

A metric which satisfies this inequality is called an ultrametric. Interpret and prove the following statement: in an ultrametric space, every triangle is isoceles.

Valuation in Geometry

Suppose A = \mathbb C[X, Y]/(Y^2 - X^3 + X), a Dedekind domain. Recall that points P on the curve Y^2 = X^3 - X correspond bijectively to maximal ideals \mathfrak m_P \subset A. Then \nu_{\mathfrak m_P}(f) is just the order of vanishing of at P.

For example, let \mathfrak m = (X, Y) = \mathfrak m_P where P = (0, 0); for convenience write \nu := \nu_{\mathfrak m}. Let us compute \nu(f) for f = X^2 - Y^4 \in A:

X^2 - Y^4 = X^2 - (X^3 - X)^2 = 2X^4 - X^6 = X^4 ( 2 - X^2).

Now \nu(2 - X^2) = 0 since f does not vanish on P. From an earlier example, we have \mathfrak m_P^2 = (X) so \nu(X) = 2 and \nu(X^2 - Y^4) = 8.

As a simple exercise, compute \nu(Y).


Composition Series

Recall in the last example here that for distinct maximal ideals \mathfrak m_1, \ldots, \mathfrak m_n of any ring A and k_1, \ldots, k_n \ge 1 we have a ring isomorphism

A/(\mathfrak m_1^{k_1} \ldots \mathfrak m_n^{k_n}) \cong A/\mathfrak m_1^{k_1} \times \ldots \times A/\mathfrak m_n^{k_n}.

This gives the following.

Proposition 3.

If A is a Dedekind domain with finitely many maximal ideals \mathfrak m_1, \ldots, \mathfrak m_n, then it is a PID.


It suffices to prove that each \mathfrak m_i is principal. Let us consider \mathfrak m_1. Since \mathfrak m_i^2 \subsetneq \mathfrak m_i we can pick y \in \mathfrak m_1 - \mathfrak m_1^2. From the isomorphism

A/(\mathfrak m_1 \ldots\mathfrak m_n) \cong A/\mathfrak m_1 \times \ldots \times A/\mathfrak m_n,

we can find x\in A such that

x\equiv y \pmod {\mathfrak m_1}, \ x \equiv 1 \pmod {\mathfrak m_2}, \ \ldots, \ x \equiv 1 \pmod {\mathfrak m_n}.

Then x is divisible by \mathfrak m_1 but not its square, and not divisible by \mathfrak m_2, \ldots, \mathfrak m_n. Thus \mathfrak m_1 = (x). ♦

Proposition 4.

Let \mathfrak a \subseteq A be a non-zero ideal; write

\mathfrak a = \mathfrak m_1^{k_1} \ldots \mathfrak m_n^{k_n}.

Then the composition factors of A/\mathfrak a as an A-module comprise of k_i copies of A/\mathfrak m_i for 1 \le i \le n. In particular

l_A(A/\mathfrak a) = \sum_{i=1}^n k_i.


Recall that the composition factors of A/\mathfrak a as an A-module are identical to those as a module over itself.


The above ring isomorphism

A/(\mathfrak m_1 \ldots\mathfrak m_n) \cong A/\mathfrak m_1 \times \ldots \times A/\mathfrak m_n

is in fact an isomorphism of A-algebras. Hence l_A(A/\mathfrak a) = \sum_i l_A(A/\mathfrak m_i^{n_i}). We obtain a series of submodules of A/\mathfrak m^n

0 \subsetneq \mathfrak m^{n-1}/\mathfrak m^n \subsetneq \mathfrak m^{n-2}/\mathfrak m^n \subsetneq \ldots \subsetneq \mathfrak m/\mathfrak m^n \subsetneq A/\mathfrak m^n

with successive quotients \mathfrak m^i/\mathfrak m^{i+1}. It suffices to show each of these is of dimension 1 over A/\mathfrak m. To prove that, we recall that A_{\mathfrak m} is a dvr so \mathfrak n := \mathfrak m A_{\mathfrak m} is a principal ideal. Since \mathfrak m^i / \mathfrak m^{i+1} is already a module over A_{\mathfrak m} we have:

\dim_{A/\mathfrak m} \mathfrak m^i/\mathfrak m^{i+1} = \dim_{A_{\mathfrak m}/\mathfrak n} \mathfrak n^i / \mathfrak n^{i+1} = 1

as desired. ♦


Consider the special case where A = k[V] is the coordinate ring of a variety V over an algebraically closed field k. If A is a Dedekind domain, then since A/\mathfrak m \cong k for each maximal ideal \mathfrak m \subset A we have

l_A(A/\mathfrak a) = \dim_k A/\mathfrak a.


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